# Finding perimeter of polygon inscribed in circle

n equally spaced points are taken on the circumference of a circle of radius 1. How do you find the perimeter of the resulting regular polygon obtained by joining the n points in order?

• Connect the vertices to the center. Now you have $n$ equal isosceles triangles. How do you find the lengths of their bases? Commented Sep 21, 2020 at 17:39

Hint:

Partition the regular polygon in $$n$$ isosceles triangles with a common vertex at the centre of the circle. The altitude of such a triangle is equal to $$\cos\frac{\pi}n$$. Can you find its base?

Answer: $$n \sqrt{2r^2 (1-\cos(2\pi/n))}$$. This formula is equivalent to:

$$2nr \sin(\pi/n)$$

Explanation: Using the cosine rule: since we are given 2 sides and the opposite angle of the missing side, we obtain the last side with the following formula: $$c^2 = a^2 + b^2 - 2ab \cos(\theta)$$ $$\implies \text{missing}^2 = 2r^2 - 2r^2 \cos(2\pi/n)$$

The equivalence to the second formula comes from the fact that $$\sqrt{2-2\cos(2x)} = 2\sin(x)$$

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Honestly, I don't know why people on this website are unable to provide quick answers to easy problems. I had this same question, and even though it only took me like 5 minutes to derive the formula, this is the kind of stuff you would actually prefer a formula answer instead of a vague hint.

Bisect one triangle of the polygon by a radial line. Angle at circle center is $$\dfrac {2\pi}{2n}.$$

We see a right angled triangle of sides:

$$( 1, \cos \frac {\pi}{n}, \sin \frac {\pi}{n})$$

Directly the polygon perimeter is $$2 n \sin \frac {\pi}{n}$$

Give it a check:

$$\lim_{~~n\to \infty} 2 n \sin \dfrac {\pi}{n} = 2\pi$$